What is the slope of the line tangent to $f(x) = -2x^{2}+2x+5$ at $x = 0$ ?
The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x+h)^{2}+2(x+h)+5) - (-2x^{2}+2x+5)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x^{2}+2x h+h^{2})+2(x+h)+5) - (-2x^{2}+2x+5)}{h}$ $ = \lim_{h \to 0} \frac{-2x^{2}-4(x h)-2h^{2}+2x+2h+5+2x^{2}-2x-5}{h}$ $ = \lim_{h \to 0} \frac{-4(x h)-2h^{2}+2h}{h}$ $ = \lim_{h \to 0} -4x-2h+2$ $ = -4x+2$ $ = (-4)(0)+2$ $ = 2$